If a Poisson variate $X$ satisfies $P(X=2) = P(X=3)$, then $P(X=5) =$

$A$ person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $X$ is observed, then the range of $X$ is

$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:

If the probability function of a random variable $X$ is given by $P(X=j) = \frac{1}{2^j}$ for $j = 1, 2, 3, \ldots, \infty$,then the variance of $X$ is:

$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cap F)$.

If the probability mass function (p.m.f.) of a discrete random variable $X$ is given by $P(X=x) = \frac{c}{x^3}$ for $x = 1, 2, 3$ and $0$ otherwise,then $E(X)$ is equal to: